JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Laws of Motion

Two blocks (

m = 0.5 kg

and

M = 4.5 kg

) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is

\frac{3}{7}

. Then the maximum horizontal force that can be applied on the block so that blocks move together is ______ N. (Round off to the Nearest Integer) [ Take g as

9.8 ms^{- 2}

]

Answer: 21

Solution

The table is frictionless, so the only horizontal force acting on the upper block is the friction exerted on it by

M

. When the two move together with acceleration

a

, that friction must supply

ma

, and it cannot exceed its limiting value

μ_{s} m g

ma \leq μ_{s} m g \Rightarrow a \leq μ_{s} g = \frac{3}{7} \times 9.8 = 4.2 m s^{- 2}

The applied force drives the whole system, so

F = (m + M) a

. At the largest permissible acceleration,

F_{m a x} = (0.5 + 4.5) \times 4.2 = 21 N

The maximum horizontal force for which the blocks still move together is

21

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Laws of Motion

Solution

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